A train travels at a certain …

Let the original average speed of the train be x km/hr.
Time taken to cover 63 km {tex} = \frac{{63}}{x}{/tex} hours
Time taken to cover 72 km when the speed is increased by 6 km/hr {tex} = \frac{{72}}{{x + 6}}{/tex} hours
By the question,we have,
{tex}\frac{{63}}{x} + \frac{{72}}{{x + 6}} = 3{/tex}
{tex} \Rightarrow \frac{{21}}{x} + \frac{{24}}{{x + 6}} = 1{/tex}
{tex} \Rightarrow \frac{{21x + 126 + 24x}}{{{x^2} + 6x}} = 1{/tex}
{tex} \Rightarrow{/tex} 45x + 126 = x² + 6x
{tex} \Rightarrow{/tex} x² - 39x - 126 = 0
{tex} \Rightarrow{/tex} x² - 42x + 3x - 126 = 0
{tex} \Rightarrow{/tex} x(x - 42) + 3(x - 42) = 0
{tex} \Rightarrow{/tex} (x - 42)(x + 3) = 0
{tex} \Rightarrow{/tex} x - 42 = 0 or x + 3 = 0
{tex} \Rightarrow{/tex} x = 42 or x = -3
Since the speed cannot be negative, {tex}x \ne -3{/tex}.
Thus, the original average speed of the train is 42 km/hr.